A) \[{{2}^{n}}\]
B) \[{{2}^{n}}\cos \frac{n\pi }{2}\]
C) \[{{2}^{n}}\sin \frac{n\pi }{2}\]
D) \[{{2}^{n/2}}\cos \frac{n\pi }{4}\]
Correct Answer: D
Solution :
Since \[{{(1+x)}^{n}}={{C}_{0}}+{{C}_{1}}x+{{C}_{2}}{{x}^{2}}+.....+{{C}_{n}}{{x}^{n}}\] Put\[x=i\], on both the sides, we get \[{{(1+i)}^{n}}=({{C}_{0}}-{{C}_{2}}+{{C}_{4}}-.....)+i({{C}_{1}}-{{C}_{3}}+{{C}_{5}}-.....)\] .....(i) Also, \[1+i=\sqrt{2}\,\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)\] in amplitude modulus form Þ\[{{(1+i)}^{n}}={{2}^{n/2}}{{\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)}^{n}}\] \[={{2}^{n/2}}\left( \cos \frac{n\pi }{4}+i\sin \frac{n\pi }{4} \right)\] ....(ii) Equating the real parts in (i) and (ii) we get, \[{{C}_{0}}-{{C}_{2}}+{{C}_{4}}-{{C}_{6}}+.....={{2}^{n/2}}\cos \frac{n\pi }{4}\]
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